Integral of f.g is zero

Question

Let $f \in L^2(\Omega)$ , I know that if $\int_{\Omega}{fg}=0$ for all $g \in C_c^{\infty}(\Omega)$ then $f=0$ almost everywhere

Now what if $g\in H^1_0(\Omega)$ ? Will I get $f=0$ everywhere??

Answer 1

According to me you can not get better than what you've already got. Indeed, since \begin{equation} C_c^\infty(\Omega)\subseteq H_0^1(\Omega)\end{equation} you obtain that $f=0$ almost everywhere. But $f\in L^2(\Omega)$ and so it is defined almost everywhere.

Trivially you can try to take the function \begin{equation}f(x)=0\quad \forall x\in\Omega\backslash K\end{equation} \begin{equation}f(x)=c\quad\quad \text{ on }\quad K\end{equation} where $K$ has measure $=0$